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xeze [42]
1 year ago
12

A rectangular prism has a length that is 3 inches shorter than the height and a width that is 5 inches longer than the height. F

ill in the blanks to create an equation for the volume of the prism as a function of the height.
Mathematics
1 answer:
LUCKY_DIMON [66]1 year ago
5 0

An equation for the volume of the prism as a function of the height is Volume = h³ + 2h² - 15h.

  • We are given a rectangular prism.
  • A rectangular prism is no different than a cuboid.
  • Let the height of the rectangular prism be "h".
  • The length of the rectangular prism is "h-3".
  • The width of the rectangular prism is "h+5".
  • The volume of the rectangular prism is the same as that of the cuboid.
  • The volume of the rectangular prism is the product of its length, its width, and its height.
  • The volume of the rectangular prism is (h - 3)*(h + 5)*h.
  • An equation for the volume of the prism as a function of the height is :
  • Volume = (h - 3)*(h + 5)*h
  • Volume = (h² + 2h - 15)*h
  • Volume = h³ + 2h² - 15h

To learn more about prism, visit :

brainly.com/question/12649592

#SPJ1

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When a figure is translated on a coordinate grid, what conclusion can you draw from the pre-image and image?
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Compute the line integral with respect to arc length of the function f(x, y, z) = xy2 along the parametrized curve that is the l
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Answer:

\displaystyle\frac{15\sqrt{3}}{4}-90\sqrt{146}

Step-by-step explanation:

The line integral with respect to arc length of the function f(x, y, z) = xy2 along the parametrized curve that is the line segment from (1, 1, 1) to (2, 2, 2) followed by the line segment from (2, 2, 2) to (−9, 6, 5) equals the sum of the line integral of f along each path separately.

Let  

C_1,C_2  

be the two paths.

Recall that if we parametrize a path C as (r_1(t),r_2(t),r_3(t)) with the parameter t varying on some interval [a,b], then the line integral with respect to arc length of a function f is

\displaystyle\int_{C}f(x,y,z)ds=\displaystyle\int_{a}^{b}f(r_1,r_2,r_3)\sqrt{(r'_1)^2+(r'_2)^2+(r'_3)^2}dt

Given any two points P, Q we can parametrize the line segment from P to Q as

r(t) = tQ + (1-t)P with 0≤ t≤ 1

The parametrization of the line segment from (1,1,1) to (2,2,2) is

r(t) = t(2,2,2) + (1-t)(1,1,1) = (1+t, 1+t, 1+t)

r'(t) = (1,1,1)

and  

\displaystyle\int_{C_1}f(x,y,z)ds=\displaystyle\int_{0}^{1}f(1+t,1+t,1+t)\sqrt{3}dt=\\\\=\sqrt{3}\displaystyle\int_{0}^{1}(1+t)(1+t)^2dt=\sqrt{3}\displaystyle\int_{0}^{1}(1+t)^3dt=\displaystyle\frac{15\sqrt{3}}{4}

The parametrization of the line segment from (2,2,2) to  

(-9,6,5) is

r(t) = t(-9,6,5) + (1-t)(2,2,2) = (2-11t, 2+4t, 2+3t)  

r'(t) = (-11,4,3)

and  

\displaystyle\int_{C_2}f(x,y,z)ds=\displaystyle\int_{0}^{1}f(2-11t,2+4t,2+3t)\sqrt{146}dt=\\\\=\sqrt{146}\displaystyle\int_{0}^{1}(2-11t)(2+4t)^2dt=-90\sqrt{146}

Hence

\displaystyle\int_{C}f(x,y,z)ds=\displaystyle\int_{C_1}f(x,y,z)ds+\displaystyle\int_{C_2}f(x,y,z)ds=\\\\=\boxed{\displaystyle\frac{15\sqrt{3}}{4}-90\sqrt{146}}

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proof

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</span>
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